Cos2x=sin (3pi/2-x) решить уравнение

Cos2x = sin (3pi/2 - x)

cos2x = - cosx

cos2x + cosx = 0

2cos^2x + cosx - 1 = 0

пусть cosx = t, t ∈ [ - 1; 1]

2t^2 + t - 1 = 0

D = 1 + 4*2 = 9

t1 = ( - 1 + 3) / 4 = 2/4 = 1/2

t2 = ( - 1 - 3) / 4 = - 4/4 = - 1

cosx = 1/2

x = ± arccos (1/2) + 2pik

x = ± pi/3 + 2pik. k ∈ Z

cosx = - 1

x = pi + 2pik. k ∈ Z

Ответ:

x = ± pi/3 + 2pik. k ∈ Z

x = pi + 2pik. k ∈ Z